Question

find the value of k such that the equations x+ky+3z=0,4x+3y+kz=0,and 2x+y+2z=0 has non trival equation​

Answer 1

Answer:

x+ky+3z=0

3x+ky−2z=0

2x+3y−4z=0 has non trivial solution then

∣

∣

∣

∣

∣

∣

∣

∣

1

3

2

k

k

3

3

−2

4

∣

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=0

4k+6−k(12+4)+3(9−2k)=0

4k+6−12k−4k+27−6k=0

18k=33

k=

6

11

Equations are 6x+11y+18z=0

18x+11y−12z=0

2x+3y−4z=0

on solving we get

x=−

2

5

; y=3; z=−1

z

2

xy

=

2(−1)

2

−15

=

2

Answer 2

Answer:

The value of k such that given equations

$x+ky+3z=0\\4x+3y+kz=0\\2x+y+2z=0$

are non trivial is  $k=\frac{9}{2}$.

Step-by-step explanation:

Step 1 of 3

Given a system of the equation are:

$x+ky+3z=0\\4x+3y+kz=0\\2x+y+2z=0$

To have a non-trivial solution for the system of equations

$a_{1}x+ b_{1}y+c_{1}=0\\a_{2}x+ b_{2}y+c_{2}=0\\a_{3}x+ b_{3y}+c_{3}=0$

$\left[\begin{array}{ccc}a_{1} &b_{1} &c_{1} \\a_{2} &b_{2} &c_{2} \\a_{3} &b_{3} &c_{3} \end{array}\right] =0$

Step 2 of 3

So, applying the formula we get

$\left[\begin{array}{ccc}1&k&3\\4&3&k\\2&1&2\end{array}\right]$

$1(6-k)-k(8-2k)+3(4-6)=0\\6-k-8k+2k^{2}+12-18=0\\2k^{2}-9k=0$

Step 3 of 3

On solving the quadratic equation we get:

$2k^{2}-9k=0\\2k^{2}=9k\\2k=9\\k=\frac{9}{2}$